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Wrote a fast solution for 3 in ipython.

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Felix Martin
2018-02-02 11:19:49 +01:00
parent 1da214e4e9
commit 53b662c3a8
1 changed files with 61 additions and 19 deletions
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ipython/EulerProblem003.ipynb
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@@ -21,9 +21,7 @@
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 1, "execution_count": 1,
"metadata": { "metadata": {},
"collapsed": false
},
"outputs": [], "outputs": [],
"source": [ "source": [
"def get_primes_smaller(number):\n", "def get_primes_smaller(number):\n",
@@ -49,9 +47,7 @@
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 2, "execution_count": 2,
"metadata": { "metadata": {},
"collapsed": false
},
"outputs": [], "outputs": [],
"source": [ "source": [
"import math\n", "import math\n",
@@ -85,10 +81,8 @@
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": 5, "execution_count": 3,
"metadata": { "metadata": {},
"collapsed": false
},
"outputs": [ "outputs": [
{ {
"name": "stdout", "name": "stdout",
@@ -103,7 +97,8 @@
" return get_prime_factors(number)[-1]\n", " return get_prime_factors(number)[-1]\n",
"\n", "\n",
"assert(get_largest_prime(13195) == 29)\n", "assert(get_largest_prime(13195) == 29)\n",
"print(get_largest_prime(600851475143))" "#print(get_largest_prime(600851475143))\n",
"print(6857) # computed the previously but remove it so that we can reexecute the complete kernel"
] ]
}, },
{ {
@@ -117,12 +112,59 @@
}, },
{ {
"cell_type": "code", "cell_type": "code",
"execution_count": null, "execution_count": 4,
"metadata": { "metadata": {},
"collapsed": true "outputs": [
}, {
"outputs": [], "name": "stdout",
"source": [] "output_type": "stream",
"text": [
"6857\n"
]
}
],
"source": [
"def is_prime(n, smaller_primes):\n",
" for s in smaller_primes:\n",
" if n % s == 0:\n",
" return False\n",
" if s * s > n:\n",
" return True\n",
" return True\n",
"\n",
"def prime_generator_function():\n",
" primes = [2, 3, 5, 7]\n",
" for p in primes:\n",
" yield p\n",
" while True:\n",
" p += 2\n",
" if is_prime(p, primes):\n",
" primes.append(p)\n",
" yield p\n",
" \n",
"def get_prime_factors(number):\n",
" prime_generator = prime_generator_function()\n",
" remainder = number\n",
" factors = []\n",
" for p in prime_generator:\n",
" while remainder % p == 0:\n",
" remainder /= p\n",
" factors.append(p)\n",
" if remainder == 1 or p * p > number:\n",
" break\n",
" if remainder != 1:\n",
" factors.append(remainder)\n",
" return factors\n",
"\n",
"print(get_largest_prime(600851475143))\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here we go. Obviously much better than precalculation primes that we never need."
]
} }
], ],
"metadata": { "metadata": {
@@ -141,9 +183,9 @@
"name": "python", "name": "python",
"nbconvert_exporter": "python", "nbconvert_exporter": "python",
"pygments_lexer": "ipython3", "pygments_lexer": "ipython3",
"version": "3.5.4" "version": "3.6.3"
} }
}, },
"nbformat": 4, "nbformat": 4,
"nbformat_minor": 0 "nbformat_minor": 1
} }